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3 years ago

How do you solve the compound inequality? 6 – x > 15 or 2x – 9 ≥ 3

Mathematics

1 answer:

Masja [62]3 years ago

4 0

Answer:

x < -9 or x ≥ 6

Step-by-step explanation:

6 – x > 15 or 2x – 9 ≥ 3

Solve each inequality

6-x > 15

Subtract 6 from each side

6-x-6 > 15-6

-x > 9

Multiply by -1 remembering to flip the inequality

x < -9

Then the other inequality

2x – 9 ≥ 3

Add 9 to each side

2x -9+9≥ 3 +9

2x≥ 12

Divide by 2

2x/2 ≥ 12/2

x ≥ 6

Put back together

x < -9 or x ≥ 6

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In the local frisbee league, teams have 7 members and each of the 4 teams takes turns hosting tournaments. At each tournament, e

mars1129 [50]

There are 4 teams in total and each team has 7 members. One of the team will be the host team.

Tournament committee will be made from 3 members from the host team and 2 members from each of the three remaining teams. Selecting the members for tournament committee is a combinations problem. We have to select 3 members out 7 for host team and 2 members out of 7 from each of the remaining 3 teams.

So total number of possible 9 member tournament committees will be equal to:

$7C3 \times 7C2 \times 7C2 \times 7C2\\ \\ = 324135$

This is the case when a host team is fixed. Since any team can be the host team, there are 4 possible ways to select a host team. So the total number of possible 9 member tournament committee will be:

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3 years ago

How to completely factor t^2+4tv+4v^2

Anna71 [15]

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3 years ago

The value of y is directly proportional to the value of x. If y = 180 when x = 50, the value of y when x = 25?

Triss [41]

Answer: y = 90

Step-by-step explanation:

Since x and y are proportional, when x =25 (which is 50 divided by 2), y would be 90 since 90 is 180 divided by 2.

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3 years ago

Read 2 more answers

Factor completely, then place the answer in the proper location on the grid.

Tresset [83]

$\mathrm{Factor\:out\:common\:term\:}9 \ \textgreater \ 9\left(x^4-25y^8\right)$

$Factor \ \textgreater \ x^4-25y^8$

$\mathrm{Apply\:difference\:of\:squares\:rule:\:}x^2-y^2=\left(x+y\right)\left(x-y\right)$
$x^2-5y^4=\left(x+\sqrt{5}y^2\right)\left(x-\sqrt{5}y^2\right)$

$x^2-5y^4 \ \textgreater \ \mathrm{Apply\:difference\:of\:squares\:rule:\:}x^2-y^2=\left(x+y\right)\left(x-y\right)$
$x^2-5y^4=\left(x+\sqrt{5}y^2\right)\left(x-\sqrt{5}y^2\right)$

$\left(x^2+5y^4\right)\left(x+\sqrt{5}y^2\right)\left(x-\sqrt{5}y^2\right)$

$9\left(x^2+5y^4\right)\left(x+\sqrt{5}y^2\right)\left(x-\sqrt{5}y^2\right)$

Hope this helps!

8 0

3 years ago

What is the value of x?<br> 5x+11°<br> 119°

frutty [35]

Answer:

21.6

Step-by-step explanation:

119 = 5x + 11

119 - 11 = 5x

108 = 5x

108÷5 = x

21.6 = x

hope this helped you.

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4 years ago